Let the given product be denoted by P. We take the logarithm on both sides using the property log(xyz) = log x + log y + log z to convert the products and powers into additive terms.

28. If $a, b, c$ are positive real numbers, then the value of $(ab)^{\log\left(\frac{a}{b}\right)} \cdot (bc)^{\log\left(\frac{b}{c}\right)} \cdot (ca)^{\log\left(\frac{c}{a}\right)}$ is (A) 0 (B) $-1$ (C) 1 (D) none of these 29. If $\log(2a - 3b) = \log a - \log b$ , then $a =$ (A) $\frac{3b^2}{2b - 1}$ (B) $\frac{3b}{2b - 1}$ (C) $\frac{b^2}{2b + 1}$ (D) $\frac{3b^2}{2b + 1}$ 30. If $\frac{\log x}{2a + 3b - 5c} = \frac{\log y}{2b + 3c - 5a} = \frac{\log z}{2c + 3a - 5b}$ , then $xyz =$ (A) 2 (B) 1 (C) 0 (D) $-1$ 31. If $\frac{\log 3}{x - y} = \frac{\log 5}{y - z} = \frac{\log 7}{z - x}$ , then $3^{x+y} \cdot 5^{y+z} \cdot 7^{z+x} =$ (A) 0 (B) 2 (C) 1 (D) none of these 32. If $\frac{\log_7 a}{2} = \frac{\log_7 b}{3} = \frac{\log_7 c}{4}$ and $a^{1/2} \, b^{1/3} \, c^{1/4} = 24$ , then (A) $a = 24$ (B) $b = 81$ (C) $c = 64$ (D) $c = 256$ 33. If $a = \log_{24} 12$ , $b = \log_{36} 24$ and $c = \log_{48} 36$ , then $abc =$ (A) $2bc - 1$ (B) $2bc + 1$ (C) $bc - 1$ (D) $bc + 1$ 34. If $\frac{\log_a x}{\log_{ab} x} = 4 + k + \log_a b$ , then $k =$ (A) 0 (B) 1 (C) $-2$ (D) $-3$ 35. If $\frac{\log a}{b - c} = \frac{\log b}{c - a} = \frac{\log c}{a - b}$ , then $a^a \, b^b \, c^c =$ (A) 0 (B) 1 (C) $abc$ (D) none of these 36. If $\log_y x = \log_z y = \log_x z$ , then (A) $x < y < z$ (B) $x > y \ge z$ (C) $x < y \le z$ (D) $x = y = z$ 37. If $x = -2$ , then the value of $\log_4 \left(\frac{x^2}{4}\right) - 2\log_4(4x^4)$ is (A) 2 (B) $-4$ (C) $-6$ (D) none 38. $\frac{\log_3 11}{\log_9 13} \cdot \frac{\log_7 11}{\log_{\sqrt{7}} 13} =$ (A) 1 (B) $-1$ (C) 0 (D) none of these 39. If $\log_e 2 \cdot \log_b 625 = \log_{10} 16 \cdot \log_e 10$ , then $b =$ (A) 4 (B) 5 (C) 1 (D) $e$ 40. If $(x - y)^2 = 121xy$ and $2\log(x + y) = k + \log x + \log y$ , then $k =$ (A) $\log 5$ (B) $\log 25$ (C) $\log 125$

Answer: (C) 1

Step-by-step solution

Step 1: Define the expression and take logarithm

Let the given product be denoted by $P$ . We take the logarithm on both sides using the property $\log\text{(xyz)} = \log x + \log y + \log z$ to convert the products and powers into additive terms.

Step 2: Apply logarithm power rule

Using the power rule of logarithms, $\log(u^v) = v \log u$ , we bring each exponent to the front as a multiplying factor.

Step 3: Expand terms using difference of squares

We rewrite the quotient and product rules: $\log(u/v) = \log u - \log v$ and $\log(uv) = \log u + \log v$ . Each term now has the form $(x - y)(x + y)$ , which simplifies using the algebraic identity to $x^2 - y^2$ .

Step 4: Sum the terms and determine the value of P

Summing all the expanded difference of squares terms cancels every square symmetrically, yielding $\log P = 0$ . Since $\log P = 0$ , the value of the original expression is $P = 1$ .

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